# Presentations of the ring of modular forms

I have seen two ways of describing the ring structure of modular (with respect to the full $SL (2, \mathbb Z)$) forms: $\mathbb C [E_4, E_6]$ and $\mathbb C [E_4,E_6, \Delta]/(E_4^3-E_6^2-1728\Delta )$, where $E_4$ and $E_6$ are the normalized Eisenstein series of weight 4 and 6, respectively.

As far as I've always known, every level 1 modular form is a polynomial in $E_4$ and $E_6$ so the first description is completely appropriate. The second one has the virtue of promoting $\Delta$ to a special place in the ring, but since $E_4$, $E_6$, and $\Delta$ do satisfy the well known relationship $E_4^3-E_6^2=1728\Delta$ nothing is really gained. So, unless I'm missing something, these two presentations are equivalent.

My question is then about a distinction that seems to be made in a paper by Sati: OP2 Bundles in M-theory. In section 3 he describes the ring of modular forms as $\mathbb Z [E_4,E_6, \Delta]/(E_4^3-E_6^2-1728\Delta )$ and he states, by results of Mahowald and Hopkins, that there are certain manifolds which map under the Witten genus to the ring $\mathbb Q [E_4, E_6]$, indicating that these are modular forms "without the discriminant."

I guess I don't really understand what this last part means. Can somebody please clarify this issue for me? Thanks in advance.

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Is there anything more sophisticated going on here beyond the fact that $1728$ is not invertible in $\mathbb{Z}$? – Qiaochu Yuan Jul 29 '11 at 2:49
I agree with Qiaochu. Notice that Sati's ring of modular forms has $\mathbb Z$ coefficients, – Grumpy Parsnip Jul 29 '11 at 3:25
I was thinking it is something along these lines.. But the fact that he switches to $\mathbb Q$ coefficients in $\mathbb Q [E_4, E_6]$ doesn't seem to go along with his comment that you get modular forms "without the discriminant." Am I still missing something? – clh Jul 29 '11 at 3:54
@clh: Dear clh, As far as I can tell, you are not missing anything (and the phrase "without the discriminant" seems misleading to me, unless its particular context gives it some other meaning which is not apparent when it is heard in isolation, as it is here). Regards, – Matt E Jul 29 '11 at 4:21
I think the phrase "without the discriminant" comes from the fact that $\mathbb Q \otimes\mathbb Z[E_4,E_6,\Delta]/(E_4^3-E_6^2-1728\Delta)\cong \mathbb Q[E_4,E_6]$, so that tensoring with $\mathbb Q$ removes the need to have the discriminant as a third generator. It's not a very good way to say it. – Grumpy Parsnip Jul 29 '11 at 14:35